Question 548: Electrostatics & Capacitance

Question

The potential at a point $x$ (measured in $\mu \,m$) due to some charges situated on the $x$-axis is given by $V\left( x \right) = 20/\left( {{x^2} - 4} \right)$ volt
The electric field $E$ at $x = 4\,\mu \,m$ is given by

Accepted Answer

Here, $V\left( x ight) = {{20} \over {{x^2} - 4}}volt$
We know that $E =  - {{dV} \over {dx}} = {d \over {dx}}\left( {{{20} \over {{x^2} - 4}}} ight)$
or, $E =  + {{40x} \over {{{\left( {{x^2} - 4} ight)}^2}}}$
At $x = 4\mu m,$
$E =  + {{40 	imes 4} \over {{{\left( {{4^2} - 4} ight)}^2}}}$
$ =  + {{160} \over {144}} =  + {{10} \over 9}volt/\mu m.$
Positive sign indicates that $\overrightarrow E $ is in $+ve$ $x$-direction.

Answer

$(10/9)$ volt / $\mu $ $m$ and in the $ + ve$ $x$ direction

Answer

$\left( {5/3} \right)$ volt/ $\mu $ $m$ and in the $-ve$ $x$ direction

Answer

$\left( {5/3} \right)$ volt/$\mu $ $m$ and in the $+ve$ $x$ direction

Answer

$\left( {10/9} \right)$ volt/ $\mu \,m$ and in the $-ve$ $x$ direction

Electrostatics & Capacitance

The potential at a point x (measured in \mu \,m) due to some charges situated on the x-axis is given by V\left( x \right) = 20/\left( {{x^2} - 4} \right) volt The electric field E at x = 4\,\mu \,m is given by

Choose the correct answer: